2021 AMC 12B Problems/Problem 12

Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$ , and $k$ be the number of elements in S.

Then, $S=x+y+z$

Firstly, when the greatest integer is removed, $\frac{S-x}{k-1}=32$

When the smallest integer is also removed, $\frac{S-x-y}{k-2}=35$

When the greatest integer is added back, $\frac{S-y}{k-1}=40$

We are given that $x=y+72$

After you substitute $x=y+72$ , you have 3 equations with 3 unknowns $S,$ , $y$ and $k$ .

$S-y-72=32k-32$

$S-2y-72=35k-70$

$S-y=40k-40$

This can be easily solved to yield $k=10$ , $y=8$ , $S=368$ .

$\therefore$ average value of all integers in the set $=S/k = 368/10 = 36.8$ , D)

~ SoySoy4444

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2021 AMC 12B Problems/Problem 12